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authorR. David Murray <rdmurray@bitdance.com>2009-11-14 22:21:32 (GMT)
committerR. David Murray <rdmurray@bitdance.com>2009-11-14 22:21:32 (GMT)
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Merged revisions 76190 via svnmerge from
svn+ssh://pythondev@svn.python.org/python/trunk ........ r76190 | r.david.murray | 2009-11-10 13:58:02 -0500 (Tue, 10 Nov 2009) | 3 lines Update the FAQ entry that explains that assignments in the local scope shadow variables in the outer scope (issue 7290). ........
-rw-r--r--Doc/faq/programming.rst101
1 files changed, 68 insertions, 33 deletions
diff --git a/Doc/faq/programming.rst b/Doc/faq/programming.rst
index faa3b80..0bc9411 100644
--- a/Doc/faq/programming.rst
+++ b/Doc/faq/programming.rst
@@ -277,39 +277,74 @@ confusing API to your users.
Core Language
=============
-How do you set a global variable in a function?
------------------------------------------------
-
-Did you do something like this? ::
-
- x = 1 # make a global
-
- def f():
- print x # try to print the global
- ...
- for j in range(100):
- if q > 3:
- x = 4
-
-Any variable assigned in a function is local to that function. unless it is
-specifically declared global. Since a value is bound to ``x`` as the last
-statement of the function body, the compiler assumes that ``x`` is
-local. Consequently the ``print x`` attempts to print an uninitialized local
-variable and will trigger a ``NameError``.
-
-The solution is to insert an explicit global declaration at the start of the
-function::
-
- def f():
- global x
- print x # try to print the global
- ...
- for j in range(100):
- if q > 3:
- x = 4
-
-In this case, all references to ``x`` are interpreted as references to the ``x``
-from the module namespace.
+Why am I getting an UnboundLocalError when the variable has a value?
+--------------------------------------------------------------------
+
+It can be a surprise to get the UnboundLocalError in previously working
+code when it is modified by adding an assignment statement somewhere in
+the body of a function.
+
+This code:
+
+ >>> x = 10
+ >>> def bar():
+ ... print(x)
+ >>> bar()
+ 10
+
+works, but this code:
+
+ >>> x = 10
+ >>> def foo():
+ ... print(x)
+ ... x += 1
+
+results in an UnboundLocalError:
+
+ >>> foo()
+ Traceback (most recent call last):
+ ...
+ UnboundLocalError: local variable 'x' referenced before assignment
+
+This is because when you make an assignment to a variable in a scope, that
+variable becomes local to that scope and shadows any similarly named variable
+in the outer scope. Since the last statement in foo assigns a new value to
+``x``, the compiler recognizes it as a local variable. Consequently when the
+earlier ``print x`` attempts to print the uninitialized local variable and
+an error results.
+
+In the example above you can access the outer scope variable by declaring it
+global:
+
+ >>> x = 10
+ >>> def foobar():
+ ... global x
+ ... print(x)
+ ... x += 1
+ >>> foobar()
+ 10
+
+This explicit declaration is required in order to remind you that (unlike the
+superficially analogous situation with class and instance variables) you are
+actually modifying the value of the variable in the outer scope:
+
+ >>> print(x)
+ 11
+
+You can do a similar thing in a nested scope using the :keyword:`nonlocal`
+keyword:
+
+ >>> def foo():
+ ... x = 10
+ ... def bar():
+ ... nonlocal x
+ ... print(x)
+ ... x += 1
+ ... bar()
+ ... print(x)
+ >>> foo()
+ 10
+ 11
What are the rules for local and global variables in Python?